Biostat 202B Homework 2
Due April 18, 2024 @ 11:59PM
Question 1
Answer:
\[\begin{align*} nS_{n}^2 &= \sum_{i=1}^{n} (X_i-\bar{X})^2 \\ &= \sum_{i=1}^{n} ((X_i-\mu)-(\bar{X}-\mu))^2 \\ &= \sum_{i=1}^{n} ((X_i-\mu)^2 - 2(X_i-\mu)(\bar{X}-\mu) + (\bar{X}-\mu)^2) \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - 2(\bar{X}-\mu)\sum_{i=1}^{n} (X_i-\mu) + n(\bar{X}-\mu)^2 \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - 2n(\bar{X}-\mu)^2 + n(\bar{X}-\mu)^2 \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - n(\bar{X}-\mu)^2 \end{align*}\]Hence, \(S_n^2=\frac{1}{n}\sum_{i=1}^{n} (X_i-\mu)^2 - (\bar{X}-\mu)^2\)
Let \(Y_i=(X_i-\mu)^2\), then \(E(Y_i)=\mu_y=\sigma^2\)
\(Var(Y_i)=E(X_i-\mu)^4-\{E(X_i-\mu)^2\}^2=E(X_i-\mu_x)^4-\sigma^4\)
Since \(\gamma = \frac{E(X_i-\mu_x)^4}{\sigma^4}\)
\(Var(Y_i)=\sigma^4(\gamma-1)\)
\(S_n^2=\frac{1}{n}\sum_{i=1}^{n} Y_i - (\bar{X}-\mu)^2=\bar{Y}-(\bar{X}-\mu)^2\)
Consider \(\sqrt{n}(S_n^2-\sigma_x^2)=\sqrt{n}\{\bar{Y}-(\bar{X}-\mu)^2-\sigma_x^2\}=\sqrt{n}\{\bar{Y}-(\bar{X}-\mu)^2-\mu_y\}=\sqrt{n}(\bar{Y}-\mu_y)-\sqrt{n}(\bar{X}-\mu)^2\)
By CLT, \(\sqrt{n}(\bar{Y}-\mu_y)\overset{D}{\rightarrow}\sigma_yZ\sim N(0,\sigma_y^2)\)
By CLT, \(\sqrt{n}(\bar{X}-\mu)\overset{D}{\rightarrow}N(0,\sigma^2)\)
By slutsky, \(\bar{X}-\mu = \frac{1}{\sqrt{n}}\sqrt{n}(\bar{X}-\mu)\overset{P}{\rightarrow}0\)
By C.M.T, \(\sqrt{n}(S_n^2-\sigma_x^2)\overset{D}{\rightarrow}N(0,\sigma^4(\gamma-1))\)
Question 2
Answer:
Since \(\frac{1}{\sqrt{n}}\rightarrow 0\) as \(n\rightarrow \infty\), we have
\(Y_n-\theta \rightarrow 0\) as \(n\rightarrow \infty\).
\(Y_n\rightarrow \theta\) in probability. (Note: There is a typo in the question)
Question 3
Answer:
By WLLN, \(\bar{X} \overset{P}{\rightarrow} p\)
\(E(X)=p\) and \(Var(X)=p(1-p)\)
By CLT,
\(\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{p(1-p)}} \overset{D}{\rightarrow} N(0,1)\)
Since \(\frac{p(1-p)}{\sqrt{\bar{X}(1-\bar{X})}}\overset{P}{\rightarrow}1\)
\(\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}} \overset{D}{\rightarrow} N(0,1)\)
\(P(|\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}}|\leq 1.96)=0.95\)
\(P(-1.96\leq \frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}}\leq 1.96)=0.95\)
\(P(\frac{-1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}\leq \bar{X}-p\leq \frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})=0.95\)
\(P(\bar{X}-\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}\leq p\leq \bar{X}+\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})=0.95\)
Hence, 95% asymptotic CI for p is \((\bar{X}-\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}, \bar{X}+\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})\)
Question 4
Answer:
Let \(N=n+m\), \(\gamma\) is Kurtosis of X
By WLLN and C.M.T,
\(\sqrt{N}\sqrt{\frac{S_X^2}{n}+\frac{S_Y^2}{m}}=\sqrt{\frac{S_X^2}{n/N}+\frac{S_Y^2}{m/N}}\overset{P}{\rightarrow}\sqrt{\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma}}\)
\(\sqrt{N}\{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)\}=\sqrt{\frac{N}{n}}\sqrt{n}(\bar{X}-\mu_X)-\sqrt{\frac{N}{m}}\sqrt{m}(\bar{Y}-\mu_Y)\)
Then,
\[\begin{align*} M(t)&=E(e^{t\sqrt{\frac{N}{n}}\sqrt{n}(\bar{X}-\mu_X)-t\sqrt{\frac{N}{m}}\sqrt{m}(\bar{Y}-\mu_Y)})\\ &=E\{e^{t\sqrt{\frac{N}{n}}\sigma_X\frac{\sqrt{n}(\bar{X}-\mu_X)}{\sigma_X}}\}E\{e^{t\sqrt{\frac{N}{m}}\sigma_Y\frac{\sqrt{m}(\bar{Y}-\mu_Y)}{\sigma_Y}}\}\\ &=M_Z(t\sqrt{\frac{1}{\gamma}}\sigma_X)M_Z(t\sqrt{\frac{1}{1-\gamma}\sigma_Y})\\ &=e^{\frac{t^2}{2}(\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma})} \text{, }\forall t \end{align*}\]Hence,
\(\sqrt{N}\{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)\}\overset{D}{\rightarrow}N(0,\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma})\)
By combining the above results,
\(\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{\sqrt{\frac{S_X^2}{n}+\frac{S_Y^2}{m}}}\overset{D}{\rightarrow}Z\sim N(0,1)\)
Question 5
(a) Answer:
Since \(X_i \sim Poisson(\theta)\),
BY WLLN,
\(\frac{\sqrt{n}(\bar{X}-\theta)}{\sqrt{\theta}} \overset{D}{\rightarrow} N(0,1)\)
\(\sqrt{n}\{\bar{X}-\theta\} \overset{D}{\rightarrow} N(0,\theta)\)
(b) Answer:
By delta method,
\(\sqrt{n}\{g(\bar{X})-g(\theta)\} \overset{D}{\rightarrow} N(0,\theta^2(g'(\theta))^2)\)
Question 6
(a) Answer:
To ensure \([g'(\theta)]^2\sigma^2(\theta)=1\), we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)
Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Exponential(\frac{1}{\theta})\)
Then, \(\sigma^2=\theta^2\)
\(g'(\theta)=\frac{1}{\sqrt{\theta^2}}=\frac{1}{\theta}\)
\(g(\theta)=\ln(\theta)\)
Hence, \(g(X_n)=\ln(X_n)\) is the variance stabilizing transformation.
(b) Answer:
Similarly, we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)
Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Poisson(\theta)\)
Then, \(\sigma^2=\theta\)
\(g'(\theta)=\frac{1}{\sqrt{\theta}}\)
\(g(\theta)=2\sqrt{\theta}\)
Hence, \(g(X_n)=2\sqrt{X_n}\) is the variance stabilizing transformation.
(c) Answer:
Similarly, we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)
Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Bernoulli(\theta)\)
Then, \(\sigma^2=\theta(1-\theta)\)
\(g'(\theta)=\frac{1}{\sqrt{\theta(1-\theta)}}\)
\(g(\theta)=\arcsin(2\theta-1)\)
Hence, \(g(X_n)=\arcsin(2X_n-1)\) is the variance stabilizing transformation.
(d) Answer:
\(\sqrt{n}(X_n-\theta) \overset{D}{\rightarrow} N(0,\sigma^2)\)
After transformation, we have
\(\sqrt{n}(2\sqrt{X_n}-2\sqrt{\theta}) \overset{D}{\rightarrow} N(0,1)\)
\(P(-1.96\leq \sqrt{n}(2\sqrt{X_n}-2\sqrt{\theta})\leq 1.96)=0.95\)
\(P(\frac{-1.96}{\sqrt{n}}\leq 2\sqrt{X_n}-2\sqrt{\theta}\leq\frac{1.96}{\sqrt{n}})=0.95\)
\(P(\frac{-1.96}{2\sqrt{n}}\leq \sqrt{X_n}-\sqrt{\theta}\leq\frac{1.96}{2\sqrt{n}})=0.95\)
\(P(\sqrt{X_n}-\frac{1.96}{2\sqrt{n}}\leq \sqrt{\theta}\leq\sqrt{X_n}+\frac{1.96}{2\sqrt{n}})=0.95\)
\(P((\sqrt{X_n}-\frac{1.96}{2\sqrt{n}})^2\leq \theta\leq(\sqrt{X_n}+\frac{1.96}{2\sqrt{n}})^2)=0.95\)